detail sap design
DESCRIPTION
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Abstract:
This project is a structural analysis and design of a new building for
Nablus Municipality. The building is composed of a reinforced concrete
structure with a steel dome and steel arches.
The approach is to do a preliminary design using 1D and 2D models.
The loads considered are gravity loads for the reinforced concrete parts and
gravity and wind loads for the steel structure parts. A 3D analysis using
SAP2000 is done and elements are designed accordingly.
The 3D analysis matches approximately the 1D analysis in most parts
of the project. Complete design is done for all structural parts, and few
structural design details are provided.
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Chapter1: Introduction
1.1 Project Description:
The project is structural analysis and design of a proposed new
building for Nablus Municipality. First a preliminary analysis and design using
1D and 2D models are made, then the analysis and design are done using 3D
model using SAP2000 program.
The building is composed of 3 stories above the ground, and an
underground parking. The area of each story is about 1580m2, and its height
is 4m.
The project has 2 structural joints which divide the building into three
parts: A,B, and C from left to right. It is composed of both reinforced concrete
and steel parts. The large dome over part B is made of steel and the vault
over the entrance is made of steel arches. The rest of the structure which
includes the two domes over the stair roofs over part B, and the dome over
minaret are made of reinforced concrete. All these descriptions are illustrated
on fig.1.1.
Fig. 1.1 3D view of municipality building
The architectural plans were designed in year 2007 by Haythem
Alsaadi, who is an architectural student at AN-Najah National University.
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1.2 Design Codes:
In this project, the American Concrete Institute (ACI) Code2005 is used in
designing concrete parts, and the British Code (BS 5950) in designing steel parts.
1.3 Materials
1. Concrete
Concrete strength for all concrete parts is B350f'c=280 Kg/cm
Modulus of elasticity equals 2.5*105 Kg/cm2.
Unit weight is 2.5 ton/m3.
2. Steel:
Modulus of elasticity equals 2.04*106 Kg/cm2.
Steel yield strength:
For steel reinforcement, is 4200 kg/cm2.
For rolled steel, is 2750 kg/cm2.
3. The unit weights of the main materials used are shown in table1.1:
Table1.1. Density of the main materials used
Density(ton/m3) Material
2.5 Reinforced concrete
1.2 Bricks
1.5 Filler
2.7 Masonry
2.5 Tiles
2.3 Mortar
2.3 plastering
1.9 Selected filler (compacted
base coarse)
0.04 Polycarbonate
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1.4 Loads:
1. Dead Loads:
Dead loads are composed of own weight of the slabs, beams, columns, walls,
Domes, and superimposed dead loads from the partitions and tiles.
o The super imposed dead loads are composed of:
Bricks, mortar and filling as shown below
Thus their weight is equal to
0.03*2.5+2.3*0.02+0.1*1.5 =0.27ton/m
partitions: consist of bricks 10cm thickness and plastering 1.5cm
from each side.
Using false ceiling height 0.45m under concrete slab, the partitions height will
equal 4-0.45-0.25=3.3m.
Average distance between partitions= 6m
(0.1*1,2*3.4+.03*2.3*3.3)/6 =0.1 ton/m
Total super imposed load on slab =0.27+0.1 =0.37ton/m2.
2. Live loads:
For the municipality building the live loads will be taken as 400 kg/m2.
3. Wind load
The Assessment of wind load according to British Standard should be made
as follows :
1. The basic wind speed V appropriate to the district where the structure is to be erected
2. The basic wind speed is multiplied by factors S1, S2 , S3 to give the design wind speed Vs for the part under consideration
Vs = V * S1 *S2 * S3
3. The design wind speed is converted to dynamic pressure q using the relationships
q = k Vs2
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4. The dynamic pressure is then multiplied by an appropriate pressure coefficient Cp to give the pressure p exerted at any point on the surface of a building
p = Cp*q Cp = Cpe -Cpi
If the value of the pressure coefficient Cp is negative this indicates that p is a
suction as distinct from a positive pressure
In Nablus the average basic wind speed is 80km/hr, at 10m above
ground in an open situation.
S1 is a topography factor, it will be taken 1 for level terrain.
S3 is a factor that takes into account of the degree of security and the
period of time in years during which there will be exposure to wind ,
normally wind loads on complete structure and buildings should be
calculated at S3 = 1 with the following exceptions :
o Temporary structures. o Structures where a shorter period of exposure to the wind
may be expected o Structures where a longer period of exposure to the wind
may be required o Structure where greater than normal safety is required
Since our building is not one of those exception then S3 = 1
k = 0.613 in SI units (N/m2 and m/s) " British code page 145".
S2 is a factor which accounts for ground roughness, building size and
height above ground. This factor is taken from Table1.2.
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Table1.2. Ground Roughness, building size and height above ground, factor S2
H
(m)
(1)open country with no
obstructions
(2) open country with
scattered windbreaks
(3)country with many
windbreaks , small
town , outskirts of
large cities
(4) surface with
large and frequent
obstructions ,city
centers
Class*
Class Class Class
A B C A B C A B C A B C
3 or
less 0.83 0.78 0.73 0.72 0.67 0.63 0.64 0.6 0.55 0.56 0.52 0.47
5 0.88 0.83 0.78 0.79 0.74 0.7 0.7 0.65 0.6 0.6 0.55 0.5
10 1 0.95 0.9 0.93 0.88 0.83 0.78 0.74 0.69 0.67 0.62 0.58
15 1.03 0.99 0.94 1 0.95 0.91 0.88 0.83 0.78 0.74 0.69 0.64
20 1.06 1.01 0.96 1.03 0.98 0.94 0.95 0.9 0.85 0.79 0.75 0.7
30 1.09 1.05 1 1.07 1.03 0.98 1.01 0.97 0.92 0.9 0.85 0.79
40 1.12 1.08 1.03 1.1 1.06 1.01 1.05 1.01 0.96 0.97 0.93 0.89
50 1.14 1.1 1.06 1.12 1.08 1.04 1.08 1.04 1 1.02 0.98 0.94
60 1.15 1.12 1.08 1.14 1.1 1.06 1.1 1.06 1.02 1.05 1.02 0.98
80 1.18 1.15 1.11 1.17 1.13 1.09 1.13 1.1 1.06 1.1 1.07 1.03
100 1.2 1.17 1.13 1.19 1.16 1.12 1.16 1.12 1.09 1.13 1.1 1.07
120 1.22 1.19 1.15 1.21 1.18 1.14 1.18 1.15 1.11 1.15 1.13 1.1
140 1.24 1.2 1.17 1.22 1.19 1.16 1.2 1.17 1.13 1.17 1.15 1.12
160 1.25 1.22 1.19 1.24 1.21 1.18 1.21 1.18 1.15 1.19 1.17 1.14
180 1.26 1.23 1.2 1.25 1.22 1.19 1.23 1.2 1.17 1.2 1.19 1.16
200 1.27 1.24 1.21 1.26 1.24 1.21 1.24 1.21 1.18 1.22 1.21 1.18
* Class A. All units of cladding, glazing and roofing and their immediate fixings
and individual members of unclad structures, class B. All buildings and
structures where neither the greatest horizontal dimension nor the greatest
vertical dimension exceeds 50m,
This building will be considered case 4, classC.
The internal pressure coefficient "Cpi" will be taken -0.3, corresponding
to case that all faces are impermeable.
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The external pressure coefficient " Cpe" will be taken from the table 1.3.
Table 1.3 Pressure coefficients Cpe for pitch roofs of rectangular clad buildings
Pressure coefficients Cpe for pitch roofs
building
height* ratio roof angle
wind angle =0
degree
wind angle =90
degree
Wind
ward
face
Leeward
face
Wind
ward
face
Leeward
face
h/w 0.5 0 -0.8 -0.4 -0.8 -0.4
5 -0.9 -0.4 -0.8 -0.4
10 -1.2 -0.4 -0.8 -0.6
20 -0.4 -0.4 -0.7 -0.6
30 0 -0.4 -0.7 -0.6
45 0.3 -0.5 -0.7 -0.6
60 0.7 -0.6 -0.7 -0.6
0.5 < h/w < 1.5 0 -0.8 -0.6 -1 -0.6
5 -0.9 -0.6 -0.9 -0.6
10 -1.1 -0.6 -0.8 -0.6
20 -0.7 -0.5 -0.8 -0.6
30 -0.2 -0.5 -0.8 -0.8
45 0.2 -0.5 -0.8 -0.8
60 0.6 -0.5 -0.8 -0.8
1.5 < h/w < b 0 -0.7 -0.6 -0.9 -0.7
5 -0.7 -0.6 -0.8 -0.8
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10 -0.7 -0.6 -0.8 -0.8
20 -0.8 -0.6 -0.8 -0.8
30 -1 -0.5 -0.8 -0.7
40 -0.2 -0.5 -0.8 -0.7
50 0.2 -0.5 -0.8 -0.7
60 0.5 -0.5 -0.8 -0.7
* h is building height, and w is building length
Wind angle will be taken zero. Building height ratio (h/w)
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Table 1.4. Wind pressure on the steel dome
point
no.
elevation
(m) S2
angle of
inclination
Cpe
(win ward
face)
Pressure
(N/m)
at the wind
ward face
Cpe
(leeward
face)
Pressure
(N/m)
at the leeward
face
1 17.32 0.668 70 0.7 135 -0.6 -66.1
2 19 0.688 60 0.7 143.3 -0.6 -43
3 20.25 0.702 40 0.2 74.57 -0.433 -19.8
4 20.91 0.708 10 1.2 -163.54 -0.4 -15.2
For Steel Arches:
The arches will be divided into 5 areas marked by 5 points, as shown on
Fig1.2. Table 1.5 gives wind pressure values calculated in a similar way as
given previously.
Fig.1.3. Steel Arches
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Table 1.5. Wind pressure on the steel arches.
point
no.
Elevation
(m) S2
Angle of
Inclination
Cpe at
windward
face
pressure
(N/m) at
windward
face
Cpe at
leeward
face
pressure
(N/m) at
leeward
face
1 12 0.645 50 0.433 92.29 -0.533 -29.34
2 13 0.668 37.6 0.355 88.46 -0.45 -20.26
3 13.75 0.688 25 -0.2 14.33 -0.4 -14.33
4 14.21 0.702 18.06 -0.555 -38.03 -0.4 -14.91
5 14.37 0.708 0 -0.8 -75.86 -0.4 -15.17
1.5 Loads combinations:
For concrete structures:
Ultimate load=1.2*DL (Dead load)+1.6*LL(live load).
For steel structures:
1. Ultimate load=1.4*DL+1.6*LL
2. Ultimate load=1.0*DL+1.4*WL(wind load)
3. Ultimate load=1.2*(DL+LL+WL)
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Chapter2: Preliminary design:
This Chapter provides manual analysis and design.
2.1 Structural Systems
As stated before, this project is divided into 3 parts( part A, B, and C
from left to right) by 2 structural joints as shown on Fig. 2.1.
The steel dome over part B is composed of hollow section steel
members covered by Polycarbonate, and the cylindrical roof is composed of
3-hinged steel I-section arches covered by purlins and polycarbonate.
All floors, except roof #3, are one way ribbed slabs on main drop
beams carried by columns, as shown on figures 2.1 through 2.6. Roof #3 is
designed as one way solid slab on main dropped beams on columns as
shown on Fig. 2.6.
Basement walls are used around the parking floor, and all the exterior
walls are composed of concrete, masonry, and blocks.
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2.2 Concrete Domes Analysis and Design :
There are 3 concrete domes, one on the minaret and two on the stair roofs.
These are the major equations of the domes :
For dead load
Meridian stress (C)
=wR(K1)/d+W(K3)/R.(a)
Hoop stress() =-wR(k2)/d-
W(K3)/Rd ..(b)
Where,
w: uniformly distributed load per
meter square on the area
W: concentrated load
R: radius of the dome
d: thickness of the dome
K1,2,3: constants which depend on the semi central angle and the angle at
which the stress is measured, are given in Table 2.1
For live and snow loads:
Meridian stress(C) =-q*R/2d.(c)
Hoop stress () =-q*R*(cos 2c)/2d.(d)
Where,
q: uniformly distributed load per square meter on horizontal projection.
c: the angle measured from the vertical to the point at which the stress is
to be measured.
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Table 2.1. K Factors
c k1 k2 k3
0 0.5 0.5
10 0.505 -0.48 5.3
20 0.516 -0.425 1.37
30 0.537 -0.33 0.64
40 0.566 -0.2 0.38
50 0.608 -0.034 0.27
51.48 0.618 0 0.26
60 0.667 0.167 0.21
70 0.747 0.402 0.18
80 0.838 0.68 0.16
90 1 1 0.16
1. Minaret Dome:
Concrete dome with ring beam below.
Dimensions:
Radius (R)=3 m, Rise = 2.22 m, c = 75
Thickness(d)= 10 cm (range 7.5cm
15cm).
Loads :
Dead load(w) =own weight=1*1*0.1*2.5=0.25 t/m
Live and snow loads(q) =0.15 t/m (on projected area).
Super imposed load, concentrated(W) =0.05 t.
Angle of super imposed load = 10 degree.
Applying Eqs. A to d, the stresses on the dome are given in Table 2.2.
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Table2.2. Hoop and Meridian stresses for minaret dome
c C dead (ton/m2)
dead (ton/m2)
C live and snow
(ton/m2)
live and snow
(ton/m2)
C total (ton/m2)
total (ton/m2)
0 4.50 -4.50 -2.25 -2.25 2.25 -6.75
10 4.55 4.32 -2.25 -0.92 2.30 3.40
20 4.10 2.96 -2.25 1.50 1.85 4.46
30 4.13 2.37 -2.25 2.14 1.88 4.51
40 4.31 1.44 -2.25 0.25 2.06 1.69
50 4.61 0.21 -2.25 -1.94 2.36 -1.73
51.48 4.68 -0.04 -2.25 1.70 2.43 1.66
60 5.04 -1.29 -2.25 -1.83 2.79 -3.12
70 5.63 -3.05 -2.25 0.45 3.38 -2.60
75 6.59 -4.58 -2.25 1.32 4.34 -3.26
All tension and compression stresses are much less than concrete
capacity, where concrete compression capacity is 2800 ton/m2, and the
tension concrete capacity is approximately equal to 280 ton/m2.
Thus only shrinkage steel is needed for both directions(meridian and
hoop).
As =0.2%*10*100 =2 cm2 use 4 8mm/m.
2. Domes on the stair roofs:
Concrete dome with ring beam below.
Dimensions:
Radius =1.8 m, Rise =0.85 m, c = 58
Thickness 10 cm (range 7.5cm 15cm).
Loads :
Dead load = only its own weight (w)=1*1*0.1*2.5=0.25 t/m.
Live and snow loads (q) =0.15 t/m (on the projected area).
Applying Eqs. A to d, the stresses on the dome are given in table 2.2.
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Table2.3. Meridian and Hoop Stresses for domes over stair roofs
c C dead (ton/m2)
dead (ton/m2)
C live and snow
(ton/m2)
live and snow
(ton/m2)
C total (ton/m2)
total (ton/m2)
0 2.25 -2.25 -1.35 -1.35 -1.35 -1.35
10 2.27 2.16 -1.35 -0.55 0.92 1.61
20 2.32 1.91 -1.35 0.90 0.97 2.81
30 2.42 1.49 -1.35 1.29 1.07 2.77
40 2.55 0.90 -1.35 0.15 1.20 1.05
50 2.74 0.15 -1.35 -1.16 1.39 -1.01
51.48 2.78 0.00 -1.35 1.02 1.43 1.02
58 3.00 -0.75 -1.35 -1.10 1.65 -1.85
All tension and compression stresses are much less than concrete
capacity, where concrete compression capacity is 2800 ton/m2, and the
tension concrete capacity is approximately equal to 280 ton/m2.
Thus only shrinkage steel is needed for both directions(meridian and
hoop).
As =0.2%*10*100 =2 cm2 use 4 8mm/m.
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2.3 Slabs Analysis and Design
Ribbed Slabs Analysis and Design:
Referring to Fig. 2.2 the thickness of slab, Hmin,,. is given using ACI Table 2.4.
Table 2.4: minimum thickness of one way solid slabs and beams
Member
Minimum thickness, h
One end
continuous
Two ends
continuous
simply
supported Cantilever
One way
solid slab 24
ln
28
ln
20
ln
10
ln
Beams 5.18
ln
21
ln
16
ln
8
ln
Hmin.= larger of ( 485/21= 23.1 cm, 360/18.5=19.5cm, 500/21=23.8 cm)=
23.8cm 25cm.
Figure below shows a proposed section of the ribbed slab:
Thus the own weight is calculated as:
Own weight= (0.08*.52+0.12*0.17)*2.5+0.4*0.17*1.2=0.2366 t/m/rib=
0.455t/m2.
Wu= 1.2*(0.455+0.27)+1.6*0.4= 1.51 t/m2=0.79t/m/rib.
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Slab1 (S1) in parts A and C :
Slab 1 load (in ton/m)
Slab 1 reaction diagram (in ton)
Slab 1 Shear force diagram (in ton)
Slab 1 bending moment diagram (in ton.m)
Max. shear at distance d =20 cm from the face of column= 1.91-
0.79*(0.15+0.2)=1.633 ton.
(Assumed columns dimensions are 30*70 cm.)
Nominal shear=1.633/0.75=2.178 ton
Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton>Vu no need for
shear reinforcement.
Max. neg. moment= 1.34 ton.m/rib
0079.280*20*12
34.1*10*61.211(*
4200
280*85.02
5
0033.014
.min yF
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As= db** = 9.120*12*0079.0 cm22 rib/12 top steel.
Min. neg. moment = 1.19 ton.m/rib
007.280*20*12
19.1*10*61.211(*
4200
280*85.02
5
As =1.675 cm2 2 12 /rib
So use 212 top steel for all spans except the exterior ends of the exterior
spans
Max. +ve moment= 0.99 ton.m/rib
0012.0)280*20*52
99.0*10*61.211(*
4200
280*85.02
5
As= 0.00127*52*20= 1.32cm2
As min.= 8.020*12*0033.0**.min dbw cm2
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Slab 1 Shear force diagram (in ton)
Slab 1 bending moment diagram (in ton.m)
Max. shear at distance d =20 cm from the face of column= 1.93-
0.79*(0.2+0.15)=1.653 ton.
(Assumed columns dimensions to be 30*70 cm.)
Nominal shear=1.653/0.75=2.2 ton
Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton>Vu no need for
shear reinforcement.
Max. neg. moment= 1.34 ton.m/rib
0079.280*20*12
34.1*10*61.211(*
4200
280*85.02
5
0033.014
.min yF
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0013.0)280*20*52
02.1*10*61.211(*
4200
280*85.02
5
As= 0.0013*52*20= 1.36cm2
As min.= 8.020*12*0033.0**.min dbw cm2
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Slab3 bending moment diagram
Max. shear at distance d =20 cm from the face of column= 2.32-
0.79(0.2+0.15)=2 ton
(Assumed columns dimensions to be 30*70 cm.)
Ultimate shear=2/0.75=2.67 ton
Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton10cm use 1 8mm stirrup/10 cm.
If Vn< Vc/2 ,no need for shear reinforcement.
Vc*0.75/2=.88 ton = 2.32-0.79*X1X1=1.82m
So at distance 0 to 1.82m (measured from left end of span 4) use 1 8mm
stirrup/10 cm, otherwise no need for shear reinforcement.
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Max. neg. moment at span 4= 1.73 ton.m /rib
01.280*20*12
73.1*10*61.211(*
4200
280*85.02
5
0033.014
.min yF
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Max. +ve moment at span 1 and 2 = 0.71 ton.m/rib
0009.0)280*20*52
71.0*10*61.211(*
4200
280*85.02
5
As= 0.0009*52*20= 0.945cm2
As min.= 8.020*12*0033.0**.min dbw cm2< 0.945cm2 .
Use rib/102 bottom steel.
For span 3 use min area of steel =.8 cm , 210 / rib.
Check if the compression depth smaller than flange thickness:
a =bf
fA
c
YS
**85.0
*
=
52*280*85.0
4200*26.2=0.77cm
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Slab Shear force diagram (ton/m)
Slab bending moment diagram (ton.m)
Max. shear at distance d =20 cm from the face of column= 1.42-
0.79*(0.2+0.15)= 1.143 ton
(Assumed columns dimensions to be 30*70 cm.)
Ultimate shear=1.243/0.75=1.524 ton
Vc= 0.53* cf *b*d*1.1=0.53* 280 *12*20*1.1*10-3=2.34ton>Vu no need for
shear reinforcement.
Max. neg. moment= 1.05 ton.m/rib
0061.280*20*12
05.1*10*61.211(*
4200
280*85.02
5
0033.014
.min yF
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00028.0)280*20*52
22.0*10*61.211(*
4200
280*85.02
5
As= 0.00028*52*20 =0.29 cm
As min.= 8.020*12*0033.0**.min dbw cm2>0.29 cm
Use rib/102 bottom steel, for all spans.
Check if the compression depth smaller than flange thickness:
a =bf
fA
c
YS
**85.0
*
=
52*280*85.0
4200*57.1=0.53cm
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Beam Load Diagram (in ton/m)
Beam Reactions Diagram (in ton)
Beam Shear Force Diagram (in ton)
Beam Bending Moment Diagram (in ton.m)
Flexure Design:
+ve moment on span 1 =13.28 ton.m
0061.0)280*45*30
28.13*10*61.211(*
4200
280*85.02
5
>0.0033
As=0.0061*45*30=8.23cm24 18
+ve moment on span 3 =10.94 ton.m
00497.0)280*45*30
94.10*10*61.211(*
4200
280*85.02
5
>0.0033
As=0.00497*45*30=6.7cm23 18
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Span 2 will be reinforced with min. bottom steel reinforcement
As min.=0.0033* 45*30 =4.45 cm use 2 18
-ve. Moment on span1 = 13.53 ton.m
0062.0)280*45*30
53.13*10*61.211(*
4200
280*85.02
5
>0.0033
As=0.0062*30*45=8.4 cm24 18
-ve. Moment on span2 = 10.75 ton.m
00488.0)280*45*30
75.10*10*61.211(*
4200
280*85.02
5
>0.0033
As=0.00488*30*45=6.59 cm23 18
Shear Design
Span1:
Shear Force at distance d=45 cm from the face of the right support =
19.45- 7.05*(0.15+0.45)=15.22 ton.
Vn= uV =
75.0
22.15=20.29 ton
Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*45= 11.97 ton.
VS=Vn-VC= 20.29-11.97=8.32 ton
dF
V
S
A
y
SV
*
VS
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S
AV =0.044, Using 8mm stirrups AV=1cm2
S=044.0
1=22.7cm=22.5cm ok. use 1 8mm stirrup/20 cm c/c.
Shear Force at distance d=45 cm from the face of the left support of
span1 = 13.69- 7.05*(0.15+0.45)=9.46 ton
Vn= uV =
75.0
46.9=12.61 ton.
Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*45= 11.97 ton.
VS=Vn-VC= 12.61-11.97=0.64ton
dF
V
S
A
y
SV
*
VS0.0034use min. shear
reinforcement.
max. spacing =min. of(d/2, 60 cm)= min. of(22.5, 60 cm)=22.5cm
S
AV =0.025, Using 8mm stirrups AV=1cm2
S=025.0
1=40cm>22.5cm ok. use 1 8mm stirrup/20 cm c/c.
If Vn< Vc/2 ,no need for shear reinforcement.
Vc*0.75/2=4.5 ton = 13.69-7.05*X1X1=1.3m
-4.5=13.69-7.05*X2X2=2.6m
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So at distance 1.3m to 2.6 m (measured from left end) use 1 8mm stirrup/40
cm.
Span2:
Use min. shear reinforcement at the ends(1 8mm stirrup/20 cm), and at
distance 1.3m to 2.6m measured from left end use 1 8mm stirrup/40 cm.
Span3:
Use min. shear reinforcement at the ends(1 8mm stirrup/20 cm), and at
distance 1.85m to 3.1m measured from left end use 1 8mm stirrup/40 cm.
Beam 3 in part B in the ground floor (BM3) :
Hmin.= cm9.4621
301015
H= 70cm, D= 65cm.
Width= 30cm.
Ultimate load on all spans =
mton /96.63.0*)25.07.0(*5.2*2.152.0
41.3
Slab 2 Load Diagram (in ton/m)
Slab1 Reaction Diagram (in ton)
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29
Beam Load Diagram (in ton/m)
Beam Reaction Diagram (in ton)
Beam Shear Force Diagram (in ton)
Beam Bending Moment Diagram (in ton.m)
Flexure Design:
+ve moment on span1,5 =21.11 ton.m
0046.0)280*65*30
11.21*10*61.211(*
4200
280*85.02
5
>0.0033
As=0.0046*65*30=8.93cm24 18
+ve moment on span 2,4 =1.88 ton.m
0003.0)280*65*30
88.1*10*61.211(*
4200
280*85.02
5
-
30
Asmin=0.0033*65*30=6.43cm23 18
+ve moment on span 3 =40 ton.m
00905.0)280*65*30
40*10*61.211(*
4200
280*85.02
5
0.0033
As=0.0043*30*65=8.44 cm24 18
-ve. Moment on span2,4 = 49.46 ton.m
011.0)280*65*30
46.49*10*61.211(*
4200
280*85.02
5
>0.0033
As=0.011*30*65=21.45 cm29 18
Shear Design
Max. shear at distance d=65 cm from the face of the support = 35.33-
6.96*(0.15+0.65)=29.76 ton.
Vn= uV =
75.0
76.29= 39.68ton
Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*65= 17.29 ton.
VS=VN-VC= 29.76-17.29=12.47 ton
dF
V
S
A
y
SV
*
VS
-
31
025.0)024.0,025.0(.max)( .min ofS
AV
-
32
Load assigned on slab 1
Slab 1 reaction diagram
Beam F1:
Hmin.= cm8.645.18
2.02.12
H= 70cm, D= 65cm.
Width= 30cm.
Ultimate load= mton /23.43.0*)25.07.0(*5.2*2.152.0
99.1
Frame loads diagram
-
33
Axial force diagram
Shear force diagram
-
34
Bending moment diagram
a. Beam flexure design:
+ve moment=49.82 ton.m
0116.0)280*65*30
82.49*10*61.211(*
4200
280*85.0
2
5
>0.0033 ok.
As=0.0116*30*65=22.62 cm28 20
-ve moment on the left end=15.98ton.m
0034.0)280*65*30
98.15*10*61.211(*
4200
280*85.0
2
5
>0.0033ok.
As=0.0034*30*65=6.7 cm24 16
-ve moment on the right end=42.72ton.m.
0097.0)280*65*30
72.42*10*61.211(*
4200
280*85.0
2
5
>0.0033ok.
As=0.0097*30*65=19 cm210 16
b. Beam shear design:
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35
Shear force at d distance from face of left column= 23.62-
4.23*(0.2+0.65)=19.6ton.
Vn= uV =
75.0
6.19=26.13 ton
Vc= 0.53* cf
*b*d= Vc= 0.53* 280 *30*65*10-3= 17.3 ton.
VS=VN-VC= 26.13-17.3=8.83 ton
dF
V
S
A
y
SV
*
VS
-
36
Shear force at d distance from face of right column= 27.99-
4.23*(0.2+0.65)=24.4ton.
Vn= uV =
75.0
4.24=32.53 ton.
Vc= 0.53* cf
*b*d= Vc= 0.53* 280 *30*65*10-3= 17.3 ton.
VS=VN-VC= 32.53-17.3=15.23 ton
dF
V
S
A
y
SV
*
VS
-
37
All the secondary beams are hidden, H=25cm, W=30cm.
Loads=own weight+ partitions weight
Loads=1.2*0.25*0.3*2.5+0.64=0.865ton/m.
Beam load diagram
Beam shear force diagram
Beam bending moment diagram
Flexure Design:
Max +ve moment on span4 =1.08 ton.m
00243.0)280*20*30
08.1*10*61.211(*
4200
280*85.0
2
5
-
38
-ve. Moment on span1 = 1.49 ton.m
0034.0)280*20*30
49.1*10*61.211(*
4200
280*85.0
2
5
>0.0033
As=0.0034*30*20=2.04 cm23 10
Use 3 10 top steel for all spans.
Use minimum bottom and top steel(3 10) for all the secondary beams.
Shear Design
Max. shear at distance d=20 cm from the face of the support = 2.09-
0.87*(0.15+0.2)=1.8 ton.
Vn= uV =
75.0
8.1= 2.4ton
Vc= 0.53* cf *b*d= Vc= 0.53* 280 *30*20= 5.32 ton.
Vn< Vc/2=2.66 ton no need for shear reinforcement.
-
39
2.5 Columns analysis and design:
B3 (in ground floor) Reaction Diagram
Design the worst column (has the max. axial load=62.06ton)
Assume column dimensions: 30cm*70cm.
Max. load on column= 62.06*4+1.2(0.3*0.65*4*2.5*4)=257.6ton
Check if r
lk u*